Question

Find the length and width of a rectangle that has the given perimeter and a maximum area. Perimeter: $P$ units.

Answer 1

Calculating the required length and width

Let $l$and $w$ denote the length and width of the rectangle.

Its perimeter is $2(l+w)$, which is given to be,$P$.

$\Rightarrow l+w=\frac{P}{2}...\left(1\right)$

Now, the area of the rectangle, is, given by, $A=lw$.

$\Rightarrow A=l(\frac{P}{2}-l)=\frac{P}{2}l-l^2$, which is a function of $l$.

We are required to maximise $A$.

We know that, for maxima of a function $A\text{'}\left(l\right)=0,$.

$$\begin{gathered} A\text{'}\left(l\right)=0 \\ \Rightarrow \frac{P}{2}-2l=0 \\ \Rightarrow l=\frac{P}{4} \end{gathered}$$.

Thus,$l=\frac{P}{4}$, gives${\mathrm{A}}_{\max }$.

Also,

$w=\frac{P}{2}-l=\frac{P}{4}units$

Hence, the required dimensions of the rectangle for the maximum area are $l=\frac{P}{4}units$, and $w=\frac{P}{4}units$.